Worked Cameron-Martin Shifts and Singular Translations

From a Path Shift to a Measure

A deterministic shift changes Brownian paths, but it need not change their probability law in a way that can be described by a density. The Cameron–Martin criterion identifies exactly which continuous shifts have this property. Its practical test is an energy calculation: the shift must be absolutely continuous, start at zero, and have a square-integrable derivative.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with its usual completed Brownian filtration. Fix $T>0$, and let $C_0([0,T])$ denote the continuous real-valued paths on $[0,T]$ that start at zero. Write $\mu$ for the law of $B$ on this path space. For a deterministic $h\in C_0([0,T])$, define its translated law by $\mu_h(A)=\mathbb{P}(B+h\in A)$ for measurable $A\subseteq C_0([0,T])$.

Definition: The Cameron–Martin space on $[0,T]$ is $H_0^1([0,T])=\{h\in C_0([0,T]): h\text{ is absolutely continuous and } \int_0^T |h'(t)|^2\,dt<\infty\}$. Its squared norm is $\|h\|_H^2=\int_0^T|h'(t)|^2\,dt$.

When $h$ belongs to this space, $h'$ is a deterministic progressively measurable integrand in $\mathcal{L}^2$: it satisfies $\mathbb{E}\int_0^T|h'(s)|^2\,ds<\infty$. Thus the Ito integral $\int_0^T h'(s)\,dB_s$ is defined. The Cameron–Martin theorem gives the density of the translated law. When $h$ is outside the space, the same theorem says the two path measures are singular, not merely that this particular density formula is unavailable.

Theorem (Cameron–Martin criterion and translation density): For $h\in C_0([0,T])$, the translated Wiener measure $\mu_h$ is equivalent to $\mu$ if and only if $h\in H_0^1([0,T])$. If $h\in H_0^1([0,T])$, then $$ \frac{d\mu_h}{d\mu}(x)= \exp\left(\int_0^T h'(s)\,dx(s)-\frac12\|h\|_H^2\right), $$ where the integral in the density is the Wiener integral, defined $\mu$-almost surely as an $L^2$ limit. If $h\notin H_0^1([0,T])$, then $\mu_h$ and $\mu$ are mutually singular.

This theorem, including the equivalence-versus-singularity dichotomy, is the standard Cameron–Martin theorem established in the earlier material on that theorem. Here the task is to apply its criterion carefully and calculate the resulting densities. The sign in the exponent corresponds to the translated law $B+h$ relative to the unshifted law $\mu$.

Linear Shifts

For $h(t)=ct$, where $c\in\mathbb{R}$, the path starts at zero and is absolutely continuous, with derivative $h'(t)=c$. Its energy is $\int_0^T c^2\,dt=c^2T$, which is finite for every real $c$. Therefore every linear shift is a Cameron–Martin shift. The integral in its density is $\int_0^T c\,dB_s=cB_T$.

Worked Example: A Linear Shift on a Fixed Horizon

Take $T=2$ and $h(t)=\tfrac32t$. Compute the Cameron–Martin norm and the density of the translated law.

The derivative is $h'(t)=\tfrac32$. Its squared norm is $\int_0^2(\tfrac32)^2\,dt=\tfrac94\cdot2=\tfrac92$. The stochastic integral is $\int_0^2\tfrac32\,dB_s=\tfrac32B_2$. Substituting both quantities in the theorem gives $$ \frac{d\mu_h}{d\mu}(B)= \exp\left(\frac32B_2-\frac12\cdot\frac92\right) =\exp\left(\frac32B_2-\frac94\right). $$

Check: Under $\mu$, $B_2$ has distribution $N(0,2)$, whose moment-generating function gives $\mathbb{E}[\exp(\tfrac32B_2)]=\exp((\tfrac32)^2\cdot2/2)=\exp(\tfrac94)$. Hence the density has expectation $\exp(\tfrac94-\tfrac94)=1$. Also, weighting the $N(0,2)$ endpoint law by this exponential shifts its mean to $\tfrac32\cdot2=3$ and leaves its variance $2$; this agrees with the endpoint $B_2+h(2)\sim N(3,2)$ under the translated law.

The normalization check is useful, but it does not replace the Cameron–Martin theorem: an endpoint density alone would not identify the law of the entire path. The theorem supplies the path-space statement. Notice also the degenerate case $c=0$: the energy is zero, the density is identically one, and the translation leaves the law unchanged.

Power-Law Shifts: The Threshold at One-Half

For real $\alpha>0$, define $h(t)=t^\alpha$ for $t>0$ and $h(0)=0$. This is continuous at zero and absolutely continuous on $[0,T]$, since its derivative $\alpha t^{\alpha-1}$ is integrable near zero. The decisive question is whether that derivative is square-integrable.

Proposition: For $\alpha>0$, the shift $h(t)=t^\alpha$ belongs to $H_0^1([0,T])$ exactly when $\alpha>\tfrac12$. In that case, $$ \|h\|_H^2=\frac{\alpha^2}{2\alpha-1}T^{2\alpha-1}. $$

Proof. For $t>0$, $h'(t)=\alpha t^{\alpha-1}$, so its squared derivative has integral $\alpha^2\int_0^T t^{2\alpha-2}\,dt$. For a real exponent $q$, $\int_0^T t^q\,dt$ is finite exactly when $q>-1$: if $q>-1$, its value is $T^{q+1}/(q+1)$; if $q=-1$, the integral from $\varepsilon$ to $T$ is $\log(T/\varepsilon)$, which diverges as $\varepsilon\downarrow0$; and if $q<-1$, the same truncated integral diverges as $\varepsilon\downarrow0$. Taking $q=2\alpha-2$, the finite-energy condition is $2\alpha-2>-1$, or $\alpha>\tfrac12$. In that case integration gives $\alpha^2T^{2\alpha-1}/(2\alpha-1)$. Since $h$ already starts at zero and is absolutely continuous, this proves both directions of the criterion. $\square$

Worked Example: A Power Shift That Has a Density

Take $T=1$, $\alpha=\tfrac34$, and $h(t)=t^{3/4}$. Find its energy and translation density, and check that the density is normalized.

The derivative is $h'(s)=\tfrac34s^{-1/4}$ for $s>0$. Its squared integral is $$ \int_0^1\left(\frac34s^{-1/4}\right)^2ds =\frac9{16}\int_0^1s^{-1/2}\,ds =\frac9{16}\cdot2=\frac98. $$ This is finite, so the shift belongs to the Cameron–Martin space. The density is therefore $$ \frac{d\mu_h}{d\mu}(B)= \exp\left(\int_0^1\frac34s^{-1/4}\,dB_s-\frac9{16}\right). $$

The deterministic integrand is progressively measurable and has expected squared-integral $\mathbb{E}\int_0^1(\tfrac34s^{-1/4})^2ds=\tfrac98$, so its Ito integral is in the prescribed $\mathcal{L}^2$ class. That integral is a centered Gaussian random variable with variance $\tfrac98$, by the Ito isometry and the Gaussian law of deterministic Brownian integrals. For a centered Gaussian $G$ of variance $v$, $\mathbb{E}[e^G]=e^{v/2}$. Here this expectation is $e^{9/16}$, so the density has expectation $e^{9/16-9/16}=1$.

Check: The norm calculation also follows from the proposition: $(\tfrac34)^2/(2\cdot\tfrac34-1)=\tfrac9{16}/\tfrac12=\tfrac98$, since $T^{2\alpha-1}=1^{1/2}=1$. Both calculations give the same energy.

Worked Example: The Boundary Power Shift Is Singular

Take $T=1$, $\alpha=\tfrac12$, and $h(t)=\sqrt{t}$. Decide whether a translation density exists and identify the measure-theoretic conclusion.

For $s>0$, $h'(s)=\tfrac12s^{-1/2}$. Its energy, truncated away from zero, is $$ \int_\varepsilon^1|h'(s)|^2\,ds =\frac14\int_\varepsilon^1\frac{1}{s}\,ds =-\frac14\log\varepsilon \longrightarrow\infty\qquad(\varepsilon\downarrow0). $$

Thus $h$ is not in the Cameron–Martin space, even though it is continuous and absolutely continuous. By the singular part of the Cameron–Martin theorem, $\mu_h$ and $\mu$ are mutually singular: there is no Radon–Nikodym density of the translated law with respect to the original Wiener measure. The failure is at the endpoint $\alpha=\tfrac12$, where the energy diverges logarithmically.

Check: The proposition's condition is strict: $\alpha>\tfrac12$, not $\alpha\ge\tfrac12$. Substituting $\alpha=\tfrac12$ into the denominator $2\alpha-1$ gives zero, consistent with the direct divergent integral.

Continuous Shifts That Are Not Absolutely Continuous

Square-integrable derivative is not the only condition: the shift must first be absolutely continuous. A standard example is the Cantor staircase $C(t)$ on $[0,1]$. It is continuous, satisfies $C(0)=0$ and $C(1)=1$, and has derivative zero almost everywhere. It is not absolutely continuous. Indeed, if it were absolutely continuous, the fundamental theorem for absolutely continuous functions would give $C(t)=C(0)+\int_0^t C'(s)\,ds=0$ for every $t$, contradicting $C(1)=1$.

Worked Example: A Cantor Staircase Translation

Let $h=C$ be the Cantor staircase on $[0,1]$. Can its translated Wiener measure be obtained by multiplying the original measure by a Cameron–Martin density?

The shift is a continuous path starting at zero, so it is a valid deterministic translation on $C_0([0,1])$. But the preceding argument shows it is not absolutely continuous, hence it is not in $H_0^1([0,1])$. The Cameron–Martin criterion therefore says that $\mu_C$ and $\mu$ are mutually singular, rather than equivalent. In particular, the formula involving $\int C'(s)\,dB_s$ cannot be used as a translation density: $C'$ is zero almost everywhere, but the path $C$ is not the zero shift.

Check: If one incorrectly substituted $C'=0$ into the density formula, one would obtain the constant $1$, which would assert $\mu_C=\mu$. That conclusion is impossible: the shift changes every path's endpoint from $B_1$ to $B_1+1$, so under $\mu_C$ the endpoint has distribution $N(1,1)$, whereas under $\mu$ it has distribution $N(0,1)$. The absolute-continuity hypothesis in the theorem prevents this invalid calculation.

Mutual singularity is a statement about measures: there is a measurable set of paths with full measure under one law and zero measure under the other. It does not mean that the paths $B$ and $B+h$ are disjoint objects, nor does it mean that the shift is discontinuous. The Cantor example is continuous; it fails the precise regularity test that defines the Cameron–Martin space.

Common Mistakes and a Practical Test

A common sign error is to write the density for the shift $B+h$ with a minus sign in front of the stochastic integral. For the linear shift $h(t)=ct$, the incorrect expression $\exp(-cB_T-\tfrac12c^2T)$ is normalized under $\mu$, so normalization alone does not catch the error. But weighting the endpoint law $N(0,T)$ by this expression gives mean $-cT$, corresponding to the shift $-ct$, not $ct$. The correct density is $\exp(cB_T-\tfrac12c^2T)$, which gives endpoint mean $cT$.

For a proposed shift, use the following order:

Check that it is continuous on $[0,T]$ and has $h(0)=0$.
Check absolute continuity; do not infer it merely from continuity or from an almost-everywhere derivative.
Compute $\int_0^T|h'(s)|^2\,ds$, including the behavior near endpoints.
If the integral is finite, insert $h'$ and the energy into the translation-density formula. If it is infinite or absolute continuity fails, the Cameron–Martin theorem gives singularity.

The drift of this reasoning is deterministic: the shift path itself is not a stochastic differential equation coefficient. The density is a random variable under the original Wiener measure, built from an Ito integral whose deterministic integrand has finite squared integral. In particular, a formal expression for the exponential is not enough; membership in the Cameron–Martin space is what licenses the formula.

Check Your Understanding

For each shift, first test continuity and absolute continuity, then apply the energy criterion where appropriate.

For $h(t)=2t$ on $[0,3]$, calculate the squared Cameron–Martin norm and write the translation density.
For $h(t)=t^{2/5}$ on $[0,1]$, determine whether the translated measure is equivalent or singular to Wiener measure.
For $h(t)=t^{3/2}$ on $[0,T]$, calculate the energy and state the density's stochastic-integral term.
Why does the fact that the Cantor staircase has derivative zero almost everywhere not imply that it is a zero Cameron–Martin shift?
For $h(t)=ct$, which sign in the exponential density produces endpoint mean $cT$ under the translated law?